12th Standard Mathematics — Complex Numbers: Online Practice Test

Take out the common factor \(i^{n}\): \(i^{n}\left(1+i+i^{2}+i^{3}\right)=i^{n}(1+i-1-i)=i^{n}\cdot 0=0\). Any four consecutive powers of \(i\) make one complete cycle \(\{1,i,-1,-i\}\) whose sum is zero, so the total is \(0\) for every \(n\).

Each term has a common factor: \(i^{n}+i^{n-1}=i^{n-1}(1+i)\). So the sum is \((1+i)\displaystyle\sum_{n=1}^{13} i^{n-1}=(1+i)\left(i^{0}+i^{1}+\cdots+i^{12}\right)\). The first twelve powers form three full cycles that cancel to zero, leaving \(i^{12}=1\). Hence the bracket equals \(1\) and the result is \((1+i)(1)=1+i\).

Multiplying by \(i\) rotates \(z\) through \(90^{\circ}\) without changing its length, so \(z\) and \(iz\) are two perpendicular sides of equal length \(|z|\) drawn from the origin. The four points \(0,\,z,\,z+iz,\,iz\) are the corners of a square of area \(|z|\cdot|iz|=|z|^{2}\). The required triangle uses three of those four corners, so it is exactly half the square: area \(=\dfrac{1}{2}|z|^{2}\).

If \(\bar z=\dfrac{1}{i-2}\), then \(z\) is the conjugate of the right-hand side. Conjugation passes through a quotient, so \(z=\overline{\left(\dfrac{1}{i-2}\right)}=\dfrac{1}{\,\overline{i-2}\,}=\dfrac{1}{-i-2}=\dfrac{-1}{i+2}\).

Modulus is multiplicative and \(|w^{n}|=|w|^{n}\), so evaluate each piece separately: \(|\sqrt{3}+i|=\sqrt{3+1}=2\), \(|4+3i|=\sqrt{16+9}=5\), and \(|8+6i|=\sqrt{64+36}=10\). Therefore \(|z|=\dfrac{2^{3}\cdot 5^{2}}{10^{2}}=\dfrac{8\cdot 25}{100}=2\).

Take the modulus of both sides. Using \(|2i|=2\), \(|z^{2}|=|z|^{2}\) and \(|\bar z|=|z|\), the equation becomes \(2|z|^{2}=|z|\). Since \(z\neq 0\) we may divide by \(|z|\), giving \(2|z|=1\), so \(|z|=\dfrac{1}{2}\).

Rewrite the condition as \(|z-(2-i)|\le 2\): the closed disc of radius \(2\) centred at \(2-i\). The point of this disc farthest from the origin lies on the line through the origin and the centre, at a distance equal to (centre distance) \(+\) (radius). Since \(|2-i|=\sqrt{4+1}=\sqrt{5}\), the maximum value of \(|z|\) is \(\sqrt{5}+2\).

By the reverse triangle inequality \(|a-b|\ge\big||a|-|b|\big|\) with \(a=z,\;b=\dfrac{3}{z}\): \(2=\left|z-\dfrac{3}{z}\right|\ge\left|\,|z|-\dfrac{3}{|z|}\,\right|\). Writing \(r=|z|\), this gives \(-2\le r-\dfrac{3}{r}\le 2\). The lower bound \(r-\dfrac{3}{r}\ge -2\) becomes (after multiplying by \(r>0\)) \(r^{2}+2r-3\ge 0\), i.e. \((r+3)(r-1)\ge 0\), forcing \(r\ge 1\). The least possible value of \(|z|\) is \(1\).

When \(|z|=1\) we have \(z\bar z=|z|^{2}=1\), so \(\bar z=\dfrac{1}{z}\). Substitute into the denominator: \(\dfrac{1+z}{1+\frac{1}{z}}=\dfrac{1+z}{\frac{z+1}{z}}=z\cdot\dfrac{1+z}{1+z}=z\).

Let \(z=x+iy\), so \(|z|=\sqrt{x^{2}+y^{2}}\) and \(|z|-z=\left(\sqrt{x^{2}+y^{2}}-x\right)-iy\). Matching imaginary parts: \(-y=2\Rightarrow y=-2\). Matching real parts: \(\sqrt{x^{2}+4}-x=1\Rightarrow\sqrt{x^{2}+4}=x+1\). Squaring, \(x^{2}+4=x^{2}+2x+1\Rightarrow 2x=3\Rightarrow x=\dfrac{3}{2}\). Hence \(z=\dfrac{3}{2}-2i\).

From \(|z_{k}|^{2}=z_{k}\bar z_{k}\) we get \(\bar z_{1}=\dfrac{1}{z_{1}},\;\bar z_{2}=\dfrac{4}{z_{2}},\;\bar z_{3}=\dfrac{9}{z_{3}}\). Factor \(z_{1}z_{2}z_{3}\) out of the given expression: \(9z_{1}z_{2}+4z_{1}z_{3}+z_{2}z_{3}=z_{1}z_{2}z_{3}\left(\dfrac{9}{z_{3}}+\dfrac{4}{z_{2}}+\dfrac{1}{z_{1}}\right)=z_{1}z_{2}z_{3}\left(\bar z_{3}+\bar z_{2}+\bar z_{1}\right)\). Taking moduli, \(12=|z_{1}||z_{2}||z_{3}|\cdot\big|\overline{z_{1}+z_{2}+z_{3}}\big|=6\,|z_{1}+z_{2}+z_{3}|\). Therefore \(|z_{1}+z_{2}+z_{3}|=2\).

A number is real exactly when it equals its own conjugate, so \(z+\dfrac{1}{z}=\bar z+\dfrac{1}{\bar z}\). Rearranging, \(z-\bar z=\dfrac{1}{\bar z}-\dfrac{1}{z}=\dfrac{z-\bar z}{z\bar z}\). Because \(z\) is not real, \(z-\bar z\neq 0\) and we cancel it to get \(1=\dfrac{1}{z\bar z}\), i.e. \(|z|^{2}=1\). Hence \(|z|=1\).

Use the identity \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=(z_{1}+z_{2}+z_{3})^{2}-2(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})\). The first bracket is \(0\). For the second, each \(|z_{k}|=1\) gives \(\bar z_{k}=\dfrac{1}{z_{k}}\), so \(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=z_{1}z_{2}z_{3}\left(\dfrac{1}{z_{3}}+\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}\right)=z_{1}z_{2}z_{3}\,\overline{(z_{1}+z_{2}+z_{3})}=0\). Both pieces vanish, so the sum of squares is \(0\).

A number \(w\) is purely imaginary when \(w+\bar w=0\). With \(w=\dfrac{z-1}{z+1}\): \(\dfrac{z-1}{z+1}+\dfrac{\bar z-1}{\bar z+1}=0\). Clearing denominators, \((z-1)(\bar z+1)+(\bar z-1)(z+1)=0\). Expanding, the cross terms cancel and we are left with \(2z\bar z-2=0\), i.e. \(|z|^{2}=1\). Hence \(|z|=1\).

\(|z+2|\) is the distance of \(z\) from \(-2\) and \(|z-2|\) is its distance from \(2\); both base points lie on the real axis. A point equidistant from \(-2\) and \(2\) sits on the perpendicular bisector of the segment joining them, which is the vertical line \(x=0\). Algebraically, \((x+2)^{2}+y^{2}=(x-2)^{2}+y^{2}\) reduces to \(8x=0\), so \(x=0\) \(-\) the imaginary axis.

First put the base in standard \(\cos\theta+i\sin\theta\) form. Since \(\sin 40^{\circ}=\cos 50^{\circ}\) and \(\cos 40^{\circ}=\sin 50^{\circ}\), the number is \(\cos 50^{\circ}+i\sin 50^{\circ}\). By De Moivre's theorem its fifth power is \(\cos 250^{\circ}+i\sin 250^{\circ}\). An argument of \(250^{\circ}\) is larger than \(180^{\circ}\), so to land in the principal range \((-180^{\circ},180^{\circ}]\) we subtract \(360^{\circ}\), giving \(-110^{\circ}\).

The point \(-1-i\) lies in the third quadrant, where both coordinates are negative. The reference angle satisfies \(\tan\alpha=\left|\dfrac{-1}{-1}\right|=1\), so \(\alpha=\dfrac{\pi}{4}\). In the third quadrant the principal argument is \(-(\pi-\alpha)=-\left(\pi-\dfrac{\pi}{4}\right)=-\dfrac{3\pi}{4}\), which lies in \((-\pi,\pi]\).

Simplify the base first: \(\dfrac{1+i}{1-i}=\dfrac{(1+i)^{2}}{(1-i)(1+i)}=\dfrac{1+2i+i^{2}}{1-i^{2}}=\dfrac{2i}{2}=i\). The condition becomes \(i^{n}=1\). Powers of \(i\) repeat with period \(4\) and equal \(1\) exactly when \(n\) is a multiple of \(4\), so the smallest positive value is \(n=4\).

Use \(1+\omega+\omega^{2}=0\), which gives \(1+\omega=-\omega^{2}\). Then \((1+\omega)^{7}=(-\omega^{2})^{7}=-\omega^{14}\). Reducing the exponent modulo \(3\) (since \(\omega^{3}=1\)) gives \(\omega^{14}=\omega^{2}\), so \((1+\omega)^{7}=-\omega^{2}\). Finally \(-\omega^{2}=1+\omega\), matching \(A+B\omega\) with \(A=1\) and \(B=1\).

Write each factor in polar form. \(1+i\sqrt{3}=2\,\operatorname{cis}\dfrac{\pi}{3}\), so its square is \(4\,\operatorname{cis}\dfrac{2\pi}{3}\). In the denominator, \(1-i\sqrt{3}=2\,\operatorname{cis}\!\left(-\dfrac{\pi}{3}\right)\) and \(4i=4\,\operatorname{cis}\dfrac{\pi}{2}\), whose product is \(8\,\operatorname{cis}\dfrac{\pi}{6}\). Dividing, the modulus is \(\dfrac{4}{8}=\dfrac{1}{2}\) and the argument is \(\dfrac{2\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{2}\). So the number is \(\dfrac{1}{2}\operatorname{cis}\dfrac{\pi}{2}\) with principal argument \(\dfrac{\pi}{2}\).

Solving \(x^{2}-x+1=0\) gives \(x=\dfrac{1\pm i\sqrt{3}}{2}=\cos 60^{\circ}\pm i\sin 60^{\circ}\), so \(\alpha\) and \(\beta\) are conjugates of modulus \(1\) with arguments \(\pm 60^{\circ}\). By De Moivre's theorem, \(\alpha^{2020}+\beta^{2020}=2\cos(2020\times 60^{\circ})\). Since \(2020\times 60^{\circ}=121200^{\circ}\equiv 240^{\circ}\pmod{360^{\circ}}\), this equals \(2\cos 240^{\circ}=2\left(-\dfrac{1}{2}\right)=-1\).

Let \(w=\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\), so \(w^{3}=\cos\pi+i\sin\pi=-1\). The four values of \(w^{3/4}\) are precisely the four fourth-roots of \(-1\). For the \(n\) distinct \(n\)th roots of a number \(a\), the product equals \((-1)^{\,n+1}a\). With \(n=4\) and \(a=-1\), the product is \((-1)^{5}(-1)=1\).

Using \(\omega^{4}=\omega\) and expanding, the determinant simplifies to \(3(\omega^{2}-\omega)\). With the standard primitive root \(\omega=\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\), we have \(\omega^{2}=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\), so \(\omega^{2}-\omega=-\sqrt{3}\,i\). Therefore the determinant equals \(3(-\sqrt{3}\,i)=-3\sqrt{3}\,i\).

Since \(i^{4n}=(i^{4})^{n}=1\), we get \(i^{4n+1}=i^{4n}\cdot i=i\) and \(i^{4n-1}=i^{4n}\cdot i^{-1}=i^{-1}=-i\). Substituting, \(\dfrac{i-(-i)}{2}=\dfrac{2i}{2}=i\).

Add all three rows into the first. Using \(1+\omega+\omega^{2}=0\), every entry of the new first row becomes \(z\), so \(z\) factors out. Two column operations then collapse the determinant to \(z^{3}\) (the leftover terms cancel using \((\omega-\omega^{2})^{2}=-3\)). The equation reduces to \(z^{3}=0\), whose only root is \(z=0\) \(-\) a single distinct value.